Fourier Heat Equation¶

$\nabla (k\nabla T) + \dot{q}_{p} = \rho c_{p} \frac{dT}{dt}$

Here, $k$ is the thermal conductivity ($W/(m·K)$), $\rho$ is the density ($kg/m^{3}$), and $c_{p}$ is the specific heat capacity ($J/(kg·K)$).

Consider the case of an infinite plate with the initial temperature $T_{0} = 300K$ immersed in a liquid that is at a different temperature $T_{f}$. In this case, the heat transfer takes place till the entire plate reaches the temperature $T_{f}$. We need to see how the temperature varies along the width of the plate, and how this temperature profile changes over time, depending on the $T_{f}$ and the property of the material, for example $c_{p}$.

Simplifying for the given case¶

Given that we have an infinite plate, only one dimension (along the thickness) is of interest to us. Further, assuming no external heat, we can put $\dot{q} = 0$. Taking the thermal conductivity to be constant. So, our equation becomes,

$k\frac{d^2T}{dx^2} = \rho c_{p} \frac{dT}{dt}$.

The initial condition is: $T(x,0) = T_{0}$. The boundary conditions are:

  1. $\frac{dT(0,t)}{dx} = 0$
  2. $-k\frac{dT(L,t)}{dx} = h[T(L,t)-T_{f}]$

For making the equation dimensionless and simplifications, we use: $Bi = \frac{hL}{k}$; $\alpha = \frac{k}{c_{p}\rho}$; $\tau = \frac{\alpha t}{L^2}$ and $X = \frac{x}{L}$

So, after separation of variables, the one-dimensional Fourier equation becomes: $\frac{1}{F}\frac{d^2F}{dX^2} = \frac{1}{G}\frac{d^2G}{d\tau^2}$

Let this be equal to $-\lambda_{n}^{2}$.

Now, we apply the boundary conditions, and solve for $\lambda_{n}$.

This simplifies to $\lambda_n tan(\lambda_{n}) = Bi$ -- (1).

For simplicity, we consider the case where $\tau > 0.2$ and we consider only the first solution of equation 1.

For a given material, changing the final Temperature $T_f$¶

In this case, $Bi$, and $\alpha$ remain unchanged.

In [4]:
fig = px.line_3d(df, x='Distance', y='Time', z='Temperature', animation_frame='Final temperature',range_x=[-5,5],range_z=[300,450], title = 'Temperature (K) profile of the plate along its thickness (m)')
fig.show()
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